Question: Is the function given below continuous/differentiable at $x=-1$ ? $g(x)=\begin{cases} -2x-1&,&x<-1 \\\\ x^2+3&,&x\geq-1 \end{cases}$ Choose 1 answer: Choose 1 answer: (Choice A) A Continuous but not differentiable (Choice B) B Differentiable but not continuous (Choice C) C Both continuous and differentiable (Choice D) D Neither continuous nor differentiable
Checking for continuity at $x=-1$ For the function to be continuous at $x=-1$, we need the two-sided limit $\lim_{x\to -1}g(x)$ to exist and be equal to $g(-1)$. This is the same as requiring that the two one-sided limits $\lim_{x\to -1^-}g(x)$ and $\lim_{x\to -1^+}g(x)$ exist and are equal to $g(-1)$. According to $g$ 's definition, $g(-1)=(-1)^2+3=4$. $\lim_{x\to -1^-}g(x)$ $-2x-1$ evaluated at $x=-1$ is equal to $1$. Since $-2x-1$ is continuous, we can be certain that $\lim_{x\to -1^-}g(x)=1$. $\lim_{x\to -1^+}g(x)$ $x^2+3$ evaluated at $x=-1$ is equal to $4$. Since $x^2+3$ is continuous, we can be certain that $\lim_{x\to -1^+}g(x)=4$. The two limits exits, but they are not equal. Therefore, the function is not continuous at $x=-1$. Graphically, the function skips a step at this point. [I would like to see that, please!] Checking for differentiability at $x=-1$ Since the function isn't continuous at $x=-1$, it cannot be differentiable at that point. In conclusion, the function is neither continuous nor differentiable at $x=-1$.